Question

A stainless steel cans of inner diameter 67 mm, thickness 3.5 mm, base thickness 5.71 mm and height 121 mm (excluding the base thickness) will be manufactured using an extrusion process. Billets used for the process have a diameter of 67 mm. The friction coefficient at the punch work piece is 0.03 and the workpiece die interface is 0.05. Assume the flow properties of the material are given as extrusion constant K=450 MPa, n = 0.16. The extrusion force is given by

• A. 4857

Answer 1

The correct option is A 4857
$\overline{F}=k{\sigma }_{avg}{A}_0$
$A_0$ is the original blank area
$ϵ=$ true strain
${\sigma }_{avg}=$mean flow stress (MPa)
where$k=0.9+1.6ϵ$
$k=0.9+1.6ϵ$
$ϵ=ln\frac{A_0}{A_f}$
$A_0=\frac{\pi }{4}\left(d_0^2\right)=\frac{\pi }{4}\left({67}^2\right)=3525.652m{m}^2$
$A_f=\frac{\pi }{4}\left[\right(74{)}^2-\left(67{)}^2\right]=775.187m{m}^2$
$ϵ=ln\frac{{67}^2}{{74}^2-{67}^2}=1.5247$
$k=0.9+1.5147\times 1.6=3.323$
${\sigma }_{avg}=\frac{450(1.5147{)}^{0.16}}{0.16+1}$
${\sigma }_{avg}=414.578MPa$
$\overline{F}=3.323\times 414.578\times \frac{\pi }{4}(67{)}^2$
$\overline{F}=4857.095kN$